# RD Sharma Solutions Class 8 Linear Equation In One Variable Exercise 9.3

## RD Sharma Solutions Class 8 Chapter 9 Exercise 9.3

Solve each of the following equations and also verify your solutions:

$Q1 \frac{2x – 3}{3x + 2} = -\frac{2}{3}$

Sol:

$\frac{2x – 3}{3x + 2} = -\frac{2}{3}$

=> 6x – 9 = -6x – 4

=> 6x + 6x = 9 – 4

=>     12x = 5

=> x =  $\frac{5}{12}$

Verification

$L.H.S = \frac{2(\frac{5}{12}) – 3}{3(\frac{5}{12}) + 2}$

$= \frac{\frac{5}{6} – 3}{\frac{5}{4} + 2}$

$= \frac{\frac{-13}{6})}{\frac{13}{4}}$

$= \frac{-4}{6}$

$= \frac{-2}{3}$

$R.H.S = \frac{-2}{3}$

Hence, L.H.S = R.H.S

$Q2. \frac{2 – y}{y + 7} = \frac{3}{5}$

Sol:

$\frac{2 – y}{y + 7} = \frac{3}{5}$

=> 10 – 5y = 3y + 21

=> 3y + 5y = 10 – 21

=>    8y = – 11

=>      y = $\frac{-11}{8}$

Verification

$L.H.S = \frac{2 – (\frac{-11}{8})}{(\frac{-11}{8}) + 7}$

$= \frac{16 + 11}{-11 + 56}$

$= \frac{27}{45}$

$= \frac{3}{5}$

$R.H.S = \frac{3}{5}$

Hence, L.H.S = R.H.S

$Q3 \frac{5x – 7}{3x} = 2$

Sol:

$\frac{5x – 7}{3x} = 2$

=> 6x = 5x – 7

=> 6x – 5x = 7

=>     x = -7

Verification

L.H.S = $\frac{5(-7) – 7}{3(-7)}$

= $\frac{-35 – 7}{-21}$

= $\frac{-42}{-21}$

= 2

R.H.S = 2

Hence, L.H.S = R.H.S

Q4 $\frac{3x + 5}{2x + 7}$ = 4

Sol:

$\frac{3x + 5}{2x + 7}$ = 4

=> 3x + 5 = 8x + 28

=> 8x – 3x = 5 – 28

=> 5x = – 23

=>     x = $\frac{-23}{5}$

Verification

L.H.S = $\frac{3(\frac{-23}{5}) + 5}{2(\frac{-23}{5}) + 7}$

= $\frac{-69 + 25}{-46 + 35}$

= $\frac{-44}{-11}$

= 4

R.H.S = 4

Hence, L.H.S = R.H.S

Q5 $\frac{2y + 5}{y + 4}$ = 1

Sol:

$\frac{2y + 5}{y + 4}$

=> 2y + 5 = y + 4

=> 2y – y = 4 – 5

=> y = -1

Verification

L.H.S = $\frac{2(-1) + 5}{(-1) + 4}$

= $\frac{-2 + 5}{3}$

= $\frac{3}{3}$

= 1

R.H.S = 1

Hence, L.H.S = R.H.S

Q6 $\frac{2x + 1}{3x – 2}$ = $\frac{5}{9}$

Sol:

$\frac{2x + 1}{3x – 2}$ = $\frac{5}{9}$

=> 18x + 9 = 15x – 10

=> 18x – 15x = -10 – 9

=>              3x = -19

=>                x = $\frac{-19}{3}$

Verification

L.H.S = $\frac{2(\frac{-19}{3}) + 1}{3(\frac{-19}{3}) – 2}$

= $\frac{-38 + 3}{-57 – 6}$

= $\frac{-35}{-63}$

= $\frac{5}{9}$

R.H.S = $\frac{5}{9}$

Hence, L.H.S = R.H.S

Q7 $\frac{1 – 9y}{19 – 3y}$ = $\frac{5}{8}$

Sol:

$\frac{1 – 9y}{19 – 3y}$ = $\frac{5}{8}$

=> 8 – 72y = 95 – 15y

=> 72y – 15y = 8 – 95

=>        57y = -87

=>            y = $\frac{-87}{57}$

=>             y = $\frac{-29}{19}$

Verification

L.H.S = $\frac{1 – 9(\frac{-29}{19})}{19 – 3(\frac{-29}{19})}$

= $\frac{19 + 261}{361 + 87}$

= $\frac{280}{448}$

= $\frac{5}{8}$

R.H.S = $\frac{5}{8}$

Hence, L.H.S = R.H.S

Q8 $\frac{2x}{3x + 1}$ = -3

Sol:

$\frac{2x}{3x + 1}$ = -3

=> 2x = -9x – 3

=>  2x + 9x = -3

=> 11x = -3

=>      x = $\frac{-3}{11}$

Verification

L.H.S = $\frac{2(\frac{-3}{11})}{3(\frac{-3}{11}) + 1}$

= $\frac{-6}{-9 + 11}$

= $\frac{-6}{2}$

= -3

R.H.S = -3

Hence, L.H.S = R.H.S

Q9 $\frac{y – (7 – 8y) }{9y – (3 + 4y)}$ = $\frac{2}{3}$

Sol:

$\frac{y – (7 – 8y) }{9y – (3 + 4y)}$ = $\frac{2}{3}$

=> $\frac{9y – 7}{5y – 3}$ = $\frac{2}{3}$

=> 27y – 21 = 10y – 6

=> 27y – 10y = 21 – 6

=>  17y = 15

=>      y = $\frac{15}{17}$

Verification

L.H.S = $\frac{9(\frac{15}{17}) – 7}{5(\frac{15}{17}) – 3}$

=  $\frac{135 – 119}{75 – 51}$

= $\frac{16}{24}$

= $\frac{2}{3}$

R.H.S = $\frac{2}{3}$

Hence, L.H.S = R.H.S

Q10 $\frac{6}{2x – 3(3 – 4x)}$ = $\frac{2}{3}$

Sol:

$\frac{6}{2x – 3(3 – 4x)}$ = $\frac{2}{3}$

=> $\frac{6}{6x – 3}$ = $\frac{2}{3}$

=> 12x – 6 = 18

=> 12x = 18 + 6

=> 12x = 24

=>      x = $\frac{24}{12}$

=>      x = 2

Verification

L.H.S = $\frac{6}{2(2) – 3(3 – 4(2))}$

= $\frac{6}{9}$

= $\frac{2}{3}$

R.H.S = $\frac{2}{3}$

Hence, L.H.S = R.H.S

Q11 $\frac{3}{2x}$$\frac{3}{2x}$ = $\frac{1}{12}$

Sol:

$\frac{3}{2x}$$\frac{3}{2x}$ = $\frac{1}{12}$

=> $\frac{4 – 9}{6x}$ = $\frac{1}{12}$

=> $\frac{-5}{x}$ = $\frac{1}{2}$

=>         x = -10

Verification

L.H. S = $\frac{3}{2(-10)}$$\frac{3}{2(10)}$

= $\frac{2}{-30}$$\frac{3}{-30}$

= $\frac{4 – 9}{-60}$

= $\frac{-5}{-60}$

= $\frac{1}{12}$

R.H.S = $\frac{1}{12}$

Hence, L.H.S = R.H.S

Q12 $\frac{3x + 5}{4x + 2}$ = $\frac{3x + 4}{4x + 7}$

Sol:

$\frac{3x + 5}{4x + 2}$ = $\frac{3x + 4}{4x + 7}$

=> $12x^{2} + 20x + 21x + 35 = 12x^{2} + 16x + 6x + 8$

=> $12x^{2} – 12x^{2} + 41x – 22x = 8 – 35$

=> 19x = -27

=>     x = $\frac{-27}{19}$

Verification

L.H.S = $\frac{3(\frac{-27}{19}) + 5}{4(\frac{-27}{19}) + 2}$

= $\frac{-81 + 95}{-108 + 38}$

= $\frac{14}{-70}$

= $\frac{-1}{5}$

R.H.S = $\frac{3(\frac{-27}{19}) + 4}{4(\frac{-27}{19}) + 7}$

= $\frac{-81 + 76}{-108 + 133}$

= $\frac{-5}{25}$

= $\frac{-1}{5}$

Hence, L.H.S = R.H.S

Q13 $\frac{7x – 2}{5x – 1}$ = $\frac{7x + }{5x + 4}$

Sol:

$\frac{7x – 2}{5x – 1}$ = $\frac{7x + }{5x + 4}$

=> $35x^{2} + 28x – 10x – 8 = 35x^{2} + 15x – 7x – 3$

=> $35x^{2} – 35x^{2} + 18x – 8x = 8 – 3$

=> 10x = 5

=>    x = $\frac{5}{10}$

= $\frac{1}{2}$

Verification

L.H.S = $\frac{7(\frac{1}{2}) – 2}{5(\frac{1}{2}) – 1}$

= $\frac{7 – 4}{5 – 2}$

= $\frac{3}{3}$

= 1

R.H.S = $\frac{7(\frac{1}{2}) + 3}{5(\frac{1}{2}) + 4}$

= $\frac{7 + 6}{5 + 8}$

= $\frac{3}{3}$

= 1

Hence, L.H.S= R.H.S

Q14 $(\frac{x + 1}{x + 2})^{2}$ = $\frac{x + 2}{x + 4}$

Sol:

$(\frac{x + 1}{x + 2})^{2}$ = $\frac{x + 2}{x + 4}$

=> $\frac{x^{2} + 2x + 1}{x^{2} + 4x + 4}$ = $\frac{x + 2}{x + 4}$

=> $x^{3} + 2x^{2} + x + 4x^{2} + 8x + 4 = x^{3} + 4x^{2} + 4x + 2x^{2} + 8x + 8$

=> $x^{3} – x^{3} + 6x^{2} – 6x^{2} + 9x – 12x = 8 – 4$

=> -3x = 4

=>     x = $\frac{-4}{3}$

Verification

L.H.S = $(\frac{\frac{-4}{3} + 1}{\frac{-4}{3} + 2})^{2}$

= $(\frac{-4 + 3}{-4 + 6})^{2}$

= $\frac{1}{4}$

R.H.S = $\frac{\frac{-4}{3} + 1}{\frac{-4}{3} + 2}$

= $\frac{-4 + 6}{-4 + 12}$

= $\frac{2}{8}$

= $\frac{1}{4}$

Hence, L.H.S = R.H.S

Q15 $(\frac{x + 1}{x – 4})^{2}$ = $\frac{x + 8}{x – 2}$

Sol:

$(\frac{x + 1}{x – 4})^{2}$ = $\frac{x + 8}{x – 2}$

=> $\frac{x^{2} + 2x + 1}{x^{2} – 8x + 16}$ = $\frac{x + 8}{x – 2}$

=> $x^{3} + 2x^{2} + x – 2x^{2} – 4x – 2 = x^{3} – 8x^{2} + 16x + 8x^{2} – 64x + 128$

=> $x^{3} – x^{3} – 3x + 48x = 128 + 2$

=> 45x = 130

=>     x = $\frac{130}{45}$

= $\frac{26}{9}$

Verification

L.H.S = $(\frac{\frac{26}{9} + 1}{\frac{26}{9} – 4})^{2}$

= $(\frac{26 + 9}{26 – 36})^{2}$

= $\frac{1225}{100}$

= $\frac{49}{4}$

R.H.S = $\frac{\frac{26}{9} + 8}{\frac{26}{9} – 2}$

= $\frac{26 + 72}{26 – 18}$

= $\frac{98}{8}$

= $\frac{49}{4}$

Hence, L.H.S = R.H.S

Q16 $\frac{9x – 7}{3x + 5}$ = $\frac{3x – 4}{x + 6}$

Sol:

$\frac{9x – 7}{3x + 5}$ = $\frac{3x – 4}{x + 6}$

=> $9x^{2} – 7x + 54x – 42 = 9x^{2} – 12x + 15x – 20$

=> $9x^{2} – 9x^{2} + 47x – 3x = –20 + 42$

=> 44x = 22

=>     x = $\frac{22}{44}$

= $\frac{1}{2}$

Verification

L.H.S = $\frac{9(\frac{1}{2}) – 7}{3\frac{1}{2} + 5}$

= $\frac{9 – 14}{3 + 10}$

= $\frac{-5}{3}$

R.H.S = $\frac{3(\frac{1}{2}) – 4}{\frac{1}{2} + 6}$

= $\frac{3 – 8}{1 + 12}$

= $\frac{-5}{13}$

Hence, L.H.S = R.H.S

Q17 $\frac{x + 2}{x + 5}$ = $\frac{x}{x + 6}$

Sol:

$\frac{x + 2}{x + 5}$ = $\frac{x}{x + 6}$

=> $x^{2} + 2x + 6x + 12 = x^{2} + 5x$

=> $x^{2} – x^{2} + 8x – 5x = -12$

=>   3x = – 12

=>      x = $\frac{-12}{3}$

=>       x = -4

Verification

L.H.S = $\frac{-4 + 2}{-4 + 5}$

= -2

R.H.S = $\frac{-4}{-4 + 6}$

= -2

Hence, L.H.S = R.H.S

Q18 $\frac{2x – (7 – 5x)}{9x – (3 + 4x) }$ = $\frac{7}{6}$

Sol:

$\frac{2x – (7 – 5x)}{9x – (3 + 4x) }$ = $\frac{7}{6}$

=> $\frac{7x – 7}{5x – 3}$ = $\frac{7}{6}$

=> 42x – 42 = 35x – 21

=> 42x – 35x = 42 – 21

=> 7x = 21

=>    x = $\frac{21}{7}$

= 3

Verification

L.H.S = $\frac{2(3) – (7 – 5(3))}{9(3) – (3 + 4(3)) }$

= $\frac{6 – (7 – 15)}{27 – (3 + 12)}$

= $\frac{6 + 8}{27 – 15}$

= $\frac{14}{12}$

= $\frac{7}{6}$

R.H.S = $\frac{7}{6}$

Hence, L.H.S = R.H.S

Q19 $\frac{15(2 – x) – 5(x + 6)}{1 – 3x}$ = 10

Sol:

$\frac{15(2 – x) – 5(x + 6)}{1 – 3x}$ = 10

=> $\frac{-20x}{1 – 3x}$ = 10

=> 10 – 30x = -20x

=> 30x – 20x = 10

=>        10x = 10

=>            x = 1

Verification

L.H.S = $\frac{15(2 – 1) – 5(1 + 6)}{1 – 3(1)}$

= $\frac{15 – 35}{-2}$

= $\frac{-20}{-2}$

= 10

R.H.S = 10

Hence, L.H.S = R.H.S

Q20  $\frac{x + 3}{x – 3}$ + $\frac{x + 2}{x – 2}$ = 2

Sol:

$\frac{x + 3}{x – 3}$ + $\frac{x + 2}{x – 2}$ = 2

=> $\frac{x + 3}{x – 3}$ = 2 – $\frac{x + 2}{x – 2}$

=> $\frac{x + 3}{x – 3}$ = $\frac{2x – 4 – x – 2}{x – 2}$

=>$\frac{x + 3}{x – 3}$ = $\frac{x – 6}{x – 2}$

=> $x^{2} – 2x + 3x – 6 = x^{2} – 3x – 6x + 18$

=> $x^{2} – x^{2} + x + 9x = 18 + 6$

=>             10x = 24

=>                 x = $\frac{24}{10}$

=>                  x = $\frac{12}{5}$

Verification

L.H.S = $\frac{\frac{12}{5} + 3}{\frac{12}{5} – 3}$ + $\frac{\frac{12}{5} + 2}{\frac{12}{5} – 2}$

= $\frac{12 + 15}{12 – 15}$ + $\frac{12 + 10}{12 – 10}$

= $\frac{27}{-3}$ + $\frac{22}{2}$

= $\frac{54 – 66}{-6}$

= $\frac{-12}{-6}$

= 2

R.H.S = 2

Hence, L.H.S = R.H.S

Q21 $\frac{(x + 2)(2x – 3) – 2x^{2} + 6}{x – 5}$ = 2

Sol:

$\frac{(x + 2)(2x – 3) – 2x^{2} + 6}{x – 5}$ = 2

=> $\frac{2x^{2} + x – 6 – 2x^{2} + 6}{x – 5}$ = 2

=> $\frac{x}{x – 5}$ = 2

=> 2x – 10 = x

=> 2x – x = 10

=>        x = 10

Verification

L.H.S = $\frac{(10 + 2)(2(10) – 3) – 2(10)^{2} + 6}{10 – 5}$

= $\frac{12(17) – 200 + 6}{5}$

= $\frac{10}{5}$

= 2

R.H.S = 2

Hence, L.H.S = R.H.S

Q22 $\frac{x^{2} – (x + 1) (x + 2)}{5x + 1}$ = 6

Sol:

$\frac{x^{2} – (x + 1) (x + 2)}{5x + 1}$ = 6

=> $\frac{x^{2} – x^{2} – 2x – x – 2}{5x + 1}$ = 6

=> $\frac{-3x – 2}{5x + 1}$ = 6

=> 30x + 6 = -3x – 2

=> 30x + 3x = -2 – 6

=>     33x = -8

=>         x = $\frac{-8}{33}$

Verification

L.H.S = $\frac{(\frac{-8}{33})^{2} – (\frac{-8}{33} + 1)(\frac{-8}{33} + 2)}{5(\frac{-8}{33}) + 1}$

= $\frac{\frac{64}{1089} – \frac{25}{33}(\frac{58}{33})}{\frac{-49}{33} + 1}$

= $\frac{\frac{64}{1089} – \frac{1450}{1089}}{\frac{-7}{33}}$

= $\frac{\frac{-1386}{1089}}{\frac{-7}{33}}$

= $\frac{42}{7}$

= 0

R.H.S = 0

Hence, L.H.S = R.H.S

Q23 $\frac{(2x + 3) – (5x – 7)}{6x + 11}$ = $\frac{-8}{3}$

Sol:

$\frac{(2x + 3) – (5x – 7)}{6x + 11}$ = $\frac{-8}{3}$

=> $\frac{-3x + 10}{6x + 11}$ = $\frac{-8}{3}$

=> -9x + 30 = -48x – 88

=> 48x – 9x = -88 – 30

=>   39x = -118

=>      x = $\frac{-118}{39}$

Verification

L.H.S = $\frac{-3(\frac{-118}{39}) + 10}{6(\frac{-118}{39}) + 11}$

= $\frac{354 + 390}{-708 + 429}$

= $\frac{744}{-279}$

= $\frac{-8}{3}$

R.H.S = $\frac{-8}{3}$

Hence, L.H.S = R.H.S

Q24 Find the positive value of x for which the given equations is satisfied

(i) $\frac{x^{2} – 9}{5 + x^{2}} = \frac{-5}{9}$

Sol:

$\frac{x^{2} – 9}{5 + x^{2}} = \frac{-5}{9}$

=> $9x^{2} – 81 = -25 – 5x^{2}$

=> $9x^{2} + 5x^{2} = 81 – 25$

=> $14x^{2} = 56$

=> $x^{2} = \frac{56}{14}$

=> $x^{2} = 4$

=>           x = 2

Verification

L.H.S = $\frac{2^{2} – 9}{5 + 2^{2}}$

= $\frac{4 – 9}{5 + 4}$

= $\frac{-5}{9}$

R.H.S = $\frac{-5}{9}$

Hence, L.H.S = R.H.S

(ii) $\frac{y^{2} + 4}{3y^{2} + 7} = \frac{1}{2}$

Sol:

$\frac{y^{2} + 4}{3y^{2} + 7} = \frac{1}{2}$

=> $3y^{2} + 7 = 2y^{2} + 8$

=> $3y^{2} – 2y^{2} = 8 – 7$

=> $y^{2} = 1$

=>  y = 1

Verification

L.H.S = $\frac{1^{2} + 4}{3(1)^{2} + 7}$

= $\frac{5}{10}$

= $\frac{1}{2}$

R.H.S = $\frac{1}{2}$

Hence, L.H.S = R.H.S

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A set of numerical facts, collected with a definite object in view, is known as